Some useful relations and identities:
[tex]\tan x=\dfrac{\sin x}{\cos x}[/tex]
[tex]\sin^2x=\dfrac{1-\cos2x}2[/tex]
[tex]\cos^2x=\dfrac{1+\cos2x}2[/tex]
By the first relation, we have
[tex]\tan^2x\sin^3x=\dfrac{\sin^5x}{\cos^2x}=\dfrac{(\sin^2x)^2\sin x}{\cos^2x}[/tex]
Applying the two latter identities, we get
[tex]\dfrac{\left(\frac{1-\cos2x}2\right)^2\sin x}{\frac{1+\cos2x}2}=\dfrac{\frac{1-2\cos2x+\cos^22x}4\sin x}{\frac{1+\cos2x}2}=\dfrac{(1-2\cos2x+\cos^22x)\sin x}{2(1+\cos2x)}[/tex]
We can apply the third identity again:
[tex]\dfrac{(1-2\cos2x+\cos^22x)\sin x}{2(1+\cos2x)}=\dfrac{\left(1-2\cos2x+\frac{1+\cos4x}2\right)\sin x}{2(1+\cos2x)}=\dfrac{(3-4\cos2x+\cos4x)\sin x}{4(1+\cos2x)}[/tex]
and this is probably as far as you have to go, but by no means is it the only possible solution.
